But it does not follow that the process _will_ go on for ever. We may come at last to a state in which both the creases are axes of symmetry at once; and then the process stops. If the paper had at first a curvilinear boundary, properly chosen, and if the axes were placed at the proper angle, it would happen that we should arrive at a {15} _regular_ curved polygon, having the two axes for axes of symmetry. The process would then stop. I will, however, suppose that the original boundary is everywhere rectilinear. It is clear then that, after every cutting, the boundary is still rectilinear. If the creases be at right angles to one another, the ultimate figure may be an irregular polygon, having its four quarters alike, such as may be inscribed in an oval; or it may have its sides so many and so small, that the ultimate appearance shall be that of an oval. But if the creases be not at right angles, the ultimate figure is a perfectly regular polygon, such as can be inscribed in a circle; or its sides may be so many and so small that the ultimate appearance shall be that of a circle. Suppose, as in MR. INGLEBY'S question, that the creases are not at right angles to each other; supposing the eye and the scissors _perfect_, the results will be as follows: First, suppose the angle made by the creases to be what the mathematicians call _incommensurable_ with the whole revolution; that is, suppose that no repetition of the angle will produce an _exact_ number of revolutions. Then the cutting will go on for ever, and the result will perpetually approach a circle. It is easily shown that no figure whatsoever, except a circle, has two axes of symmetry which make an angle incommensurable with the whole revolution. Secondly, suppose the angle of the creases commensurable with the revolution. Find out the smallest number of times which the angle must be repeated to give an exact number of revolutions. If that number be even, it is the number of sides of the ultimate polygon: if that number be odd, it is the half of the number of sides of the ultimate polygon. Thus, the paper on which I write, the whole sheet being taken, and the creases made by joining opposite corners, happens to give the angle of the creases very close to three-fourteenths of a revolution; so that fourteen repetitions of the angle is the lowest number which give an exact number of revolutions; and a very few cuttings lead to a regular polygon of fourteen sides. But if four-seventeenths of a revolution had been taken for the angle of the creases, the ultimate polygon would have had thirty-four sides. In an angle taken at hazard the chances are that the number of ultimate sides will be large enough to present a circular appearance. Any reader who chooses may amuse himself by trying results from three or more axes, whether all passing through one point or not. A. DE MORGAN. * * * * * THE BLACK-GUARD. (Vol. viii., p. 414.)